Correct answer: e41...K:b4[a]/B:b6[d] 2.Be7#[A] 1...K:b6/[б]//B:b4[c] 2.Bd4#[Б] 1.Kc7? (2.Be7#[A]/Nd3#[C]) 1...K:b4[a] 2.Be7#[A] 1...B:b4[c] 2.B